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**Unformatted text preview: **Chapter 1 Section 1.1 1. –4 –2 2 4 y(t) –4 –2 2 4 t For y > 3 / 2, the slopes are negative, and, therefore the solutions decrease. For y < 3 / 2, the slopes are positive, and, therefore, the solutions increase. As a result, y → 3 / 2 as t → ∞ 2. –4 –2 2 4 y(t) –4 –2 2 4 t For y > 3 / 2, the slopes are positive, and, therefore the solutions increase. For y < 3 / 2, the slopes are negative, and, therefore, the solutions decrease. As a result, y diverges from 3 / 2 as t → ∞ 3. 1 –4 –2 2 4 y(t) –4 –2 2 4 t For y >- 3 / 2, the slopes are positive, and, therefore the solutions increase. For y <- 3 / 2, the slopes are negative, and, therefore, the solutions decrease. As a result, y diverges from- 3 / 2 as t → ∞ 4. –4 –2 2 4 y(t) –4 –2 2 4 t For y >- 1 / 2, the slopes are negative, and, therefore the solutions decrease. For y <- 1 / 2, the slopes are positive, and, therefore, the solutions increase. As a result, y → - 1 / 2 as t → ∞ 5. –4 –2 2 4 y(t) –4 –2 2 4 t 2 For y >- 1 / 2, the slopes are positive, and, therefore the solutions increase. For y <- 1 / 2, the slopes are negative, and, therefore, the solutions decrease. As a result, y diverges from- 1 / 2 as t → ∞ 6. –4 –2 2 4 y(t) –4 –2 2 4 t For y >- 2, the slopes are positive, and, therefore the solutions increase. For y <- 2, the slopes are negative, and, therefore, the solutions decrease. As a result, y diverges from- 2 as t → ∞ 7. For the solutions to satisfy y → 3 as t → ∞ , we need y < 0 for y > 3 and y > 0 for y < 3. The equation y = 3- y satisfies these conditions. 8. For the solutions to satisfy y → 2 / 3 as t → ∞ , we need y < 0 for y > 2 / 3 and y > 0 for y < 2 / 3. The equation y = 2- 3 y satisfies these conditions. 9. For the solutions to satisfy y diverges from 2, we need y > 0 for y > 2 and y < 0 for y < 2. The equation y = y- 2 satisfies these conditions. 10. For the solutions to satisfy y diverges from 1 / 3, we need y > 0 for y > 1 / 3 and y < for y < 1 / 3. The equation y = 3 y- 1 satisfies these conditions. 11. –4 –2 2 4 y(t) –4 –2 2 4 t 3 y = 0 and y = 4 are equilibrium solutions; y → 4 if initial value is positive; y diverges from 0 if initial value is negative. 12. –4 –2 2 4 6 y(t) –4 –2 2 4 t y = 0 and y = 5 are equilibrium solutions; y diverges from 5 if initial value is greater than 5; y → 0 if initial value is less than 5. 13. –4 –2 2 4 y(t) –4 –2 2 4 t y = 0 is equilibrium solution; y → 0 if initial value is negative; y diverges from 0 if initial value is positive. 14. 4 –4 –2 2 4 y(t) –4 –2 2 4 t y = 0 and y = 2 are equilibrium solutions; y diverges from 0 if initial value is negative; y → 2 if initial value is between 0 and 2; y diverges from 2 if initial value is greater than 2....

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